Ask Question
2 August, 01:16

Your pool is 4m wide and 6m long. At the beginning the depth of the water in it was 2m, than you added 100m3 of water. How deep is the water in your pool now?

+3
Answers (2)
  1. 2 August, 01:24
    0
    6.1667 meters

    Step-by-step explanation:

    If the inicial depth of water was 2 meters, we can calculate the inicial volume of water. The volume of water in the pool can be calculated multiplying the length, width and height:

    V = 6 * 4 * 2 = 48 m3

    If you put 100m3 of water, the total volume of water will be:

    100 + 48 = 148 m3 of water.

    Now, we can find the depth (or height) of water using the same equation we used to find the volume:

    148 = 6 * 4 * h

    h = 148/24 = 6.1667 meters
  2. 2 August, 01:35
    0
    The new height of the water is 6.167 m

    Step-by-step explanation:

    The volume of the pool is given by the product of it's three dimensions width, length and height. In the initial situation we have a pool with width equals 4m, the length equals 6m and height of the water of 2m. So we can find the volume as seen bellow:

    initial volume = length*width*height = 6*4*2 = 48 m^3

    Then 100 m^3 were added to the pool, so it's new volume will be the sum of the initial volume and the added water so we have:

    final volume = initial volume + 100 = 48 + 100 = 148 m^3

    Since the only dimension that can change in this situation is the height ofthe water we can use the formula for the volume to find the new height, as seen below:

    final volume = length*width * (new height)

    148 = 6*4 * (new height)

    6*4 * (new height) = 148

    24 * (new height) = 148

    new height = 148/24 = 6.167 m

    The new height of the water is 6.167 m
Know the Answer?
Not Sure About the Answer?
Find an answer to your question 👍 “Your pool is 4m wide and 6m long. At the beginning the depth of the water in it was 2m, than you added 100m3 of water. How deep is the ...” in 📗 Mathematics if the answers seem to be not correct or there’s no answer. Try a smart search to find answers to similar questions.
Search for Other Answers